Simplicial Structures in Topology

20 I Fundamental Concepts
We wish to prove that f is constant on every open set of U. Indeed, let U ∈ U and
x ∈ U be given and suppose that f (x)=0; then [0,x] is covered by a finite subcovering
V ⊂ U. But for every y ∈ U, the interval [0,y] is covered by a finite family
of open intervals of U (that is to say, by V ∪{U}) meaning thereby that f (y)=0.
This shows that f equals either 0 or 1 on the entire open interval U. It follows that
f is continuous; moreover, since I =[0,1] is connected, by Theorem (I.1.12), f is
constant; since f (0)=0, we have f (1)=0, in other words, I is covered by a finite
subcovering of U. �
We note that, since every closed interval [a,b] where a < b (bounded) is homeomorphic
to the unit interval [0,1], every bounded and closed interval of R is compact.
We wish to prove that a finite product of compact spaces is a compact space;
this result is an immediate consequence of the following theorem:
(I.1.31) Theorem. Let B and C be compact subspaces of the topological spaces X
and Y, respectively, and let U = {A j | j ∈ J} be an open covering of B × C. Then
there are two open sets U ⊂ X and V ⊂ Y such that
B ×C ⊂ U ×V ⊂ X ×Y
and U ×V is covered by a finite number of elements of U.
Proof. The method for proving this theorem is the one used in Theorem (I.1.29) to
show the existence of two open sets U and V which contain f −1 (y1) and f −1 (y2),
respectively. We leave to the reader the task of supplying the details needed for this
proof. �
(I.1.32) Corollary. Let X1,...,Xn be compact spaces; then X1 ×...×Xn is compact.
Proof. If n = 2 we have the previous theorem where B = X = X1 and C = Y = X2;
thereafter, we proceed by induction. �
(I.1.33) Corollary. Let B and C be compact subspaces of the topological spaces X
and Y respectively. Let W be open in X ×Y such that B ×C ⊂ W . Then there are
open sets U ⊂ X and V ⊂ Y such that
B ×C ⊂ U ×V ⊂ W.
Proof. It is enough to set U = {W} in Theorem (I.1.31). �
(I.1.34) Corollary. Let X be a Hausdorff space and let B and C be disjoint compact
subspaces of X. Then there are two disjoint open sets U and V of X with B ⊂ U and
C ⊂ V.
Proof. We begin by noting that if
ΔX = {(x,x) ∈ X × X}